Question

math help ss below

Answer 1

@smokeybrown

Answer 2

For April and May we have 2 options.
For the person in between we have 6C1 = 6 options
For the rest 5 we have 5! = 120 options
Multiply all and see if its correct

Answer 3

hmm i was working a different logic, but the answer appears to be the same -- if you are confused about what its said in the above reply here is an explanation:

firstly, since this is a table of 8 people, 5 of them are randoms and 3 of them are april, may, and the person in between them

so lets do each one of them individually

for the guy in the middle, since 3 seats are always occupied by the ones on his isde we do \( \left(\begin{array}{l} (8-2) \\ \ \ \ \ \ 1 \end{array}\right) \implies \left(\begin{array}{l} 6 \\ 1 \end{array}\right) \)

lets say that may took a random seat -- april has to sit either 2 seats on the right of may, or the left of may again; makes sense? as there is that person in the middle -- knowing that we can say that there are 2 different ways may and april can sit .

lastly the random people -- its 5 of them just switching around so it would be factorial 5.

just for instance, this is how you work binomial coefficients:
\[ \left(\begin{array}{l} n \\ k \end{array}\right)=\frac{n !}{k ! \cdot(n-k) !} \]
so yeah, from here treat it like independent probability

Answer 4

Ah i already figured it out the answer was 1440

Answer 5

LOL that on the screenshot is exactly what i was thinking